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Question

# Two capacitors of capacities 1 μF and C μF are connected in series and this combination is charged to a potential difference of 120 V. If the charge on the 1 μF capacitor is 80 μC, then the energy stored in the capacitor C (in micro joule) is

A
1600
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B
1800
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C
2000
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D
1400
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Solution

## The correct option is A 1600For first capacitor C1=1 μF q1=80 μC Potential across 1μF capacitor is V1=q1C1=801 V1=80 V For second capacitor C2=C μF q2=80 μC Potential across CμF capacitor is V2=V−V1 =120−80 V2=40 V C=q2V2 C=80×10−640 C=2×10−6F Energy stored in capacitor (C μF) is U=12CV22 =12(2×10−6)×40×40 =1600×10−6J=1600 μJ

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